Proving that particular ideal is the kernel of a homomorphism of polynomial rings

Question

I have the homomorphism $f:\mathbb{R}[x,y,z]\rightarrow\mathbb{R}[t]$ with $f(x)=t,f(y)=t^2,f(z)=1$, and I want to prove that $\ker f=(x^2-y,z-1)$.

It is obvious that this ideal is in kernel, but how do I prove other way around, that every polynomial in the kernel is also in the given ideal? (usually in the books there is no explanation, they just use it as a fact)

Answer 1

We have

$$\frac{\mathbb{R}[x,y,z]}{\rm{Ker(f)}}\cong \mathbb{R}[t]$$

and $$ \frac{\mathbb{R}[x,y,z]}{(x^2-y,z-1)}\cong \frac{\mathbb{R}[x,y]}{(x^2-y)}\cong \mathbb{R}[x,x^2]\cong \mathbb{R}[x]$$

So...

Answer 2

In order to prove $\ker f\subseteq(x^2-y,z-1)$ you have to show the following:

If $p\in\mathbb R[x,y,z]$ is such that $p(t,t^2,1)=0$, then $p\in(x^2-y,z-1)$.

First write $p(x,y,z)=(z-1)q(x,y,z)+r(x,y)$. (We must have $\deg_zr<1$, that's why $r$ is a polynomial only in $x,y$.) Since $p(t,t^2,1)=0$ we get $r(t,t^2)=0$.
Now write $r(x,y)=(y-x^2)s(x,y)+w(x)$. (Analogously, $\deg_yw<1$, so $w$ is a polynomial only in $x$.) Then $r(t,t^2)=0$ implies $w(t)=0$, so $w=0$ and we are done.